在python中创建一个匿名类实例

Question

有时我需要在python中创建一个匿名类实例，就像c#:

var o = new {attr1="something", attr2=344};

但是在 python 中我是这样做的：

class Dummy: pass
o = Dummy()
o.attr1 = 'something'
o.attr2 = 344
# EDIT 1
print o.attr1, o.attr2

我怎样才能用 Pythonic 的方式和一条语句做到这一点？

Answer 1

o = type('Dummy', (object,), { "attr1": "somehing", "attr2": 344 })

o.attr3 = "test"
print o.attr1, o.attr2, o.attr3

Answer 2

如果您使用的是 Python 3.3 或更高版本，您可以使用 types.SimpleNamespace:

from types import SimpleNamespace

o = SimpleNamespace(attr1="something", attr2=344)
print(o)

# namespace(attr1='something', attr2=344)

Answer 3

类型

虽然这不完全是一个单一的陈述，但我认为围绕已接受答案的魔力创建一个包装器可以使其更具可读性。

import inspect 

# wrap the type call around a function 
# use kwargs to allow named function arguments
def create_type(name, **kwargs):
    return type(name, (object,), kwargs)

# example call to make a structure
p = create_type('foobar', xxx='barfoo', seti=0)

assert p.xxx == 'barfoo'
assert p.seti == 0

print inspect.getmembers(p)

输出

[('__class__', <type 'type'>),
 ('__delattr__', <slot wrapper '__delattr__' of 'object' objects>),
 ('__dict__', <dictproxy object at 0x9a5050>),
 ('__doc__', None),
 ('__format__', <method '__format__' of 'object' objects>),
 ('__getattribute__', <slot wrapper '__getattribute__' of 'object' objects>),
 ('__hash__', <slot wrapper '__hash__' of 'object' objects>),
 ('__init__', <slot wrapper '__init__' of 'object' objects>),
 ('__module__', '__main__'),
 ('__new__', <built-in method __new__ of type object at 0x399c578460>),
 ('__reduce__', <method '__reduce__' of 'object' objects>),
 ('__reduce_ex__', <method '__reduce_ex__' of 'object' objects>),
 ('__repr__', <slot wrapper '__repr__' of 'object' objects>),
 ('__setattr__', <slot wrapper '__setattr__' of 'object' objects>),
 ('__sizeof__', <method '__sizeof__' of 'object' objects>),
 ('__str__', <slot wrapper '__str__' of 'object' objects>),
 ('__subclasshook__', <built-in method __subclasshook__ of type object at 0x919370>),
 ('__weakref__', <attribute '__weakref__' of 'foobar' objects>),
 # here they are
 ('seti', 0),
 ('xxx', 'barfoo')]

命名元组

from collections import namedtuple

d = { 'a' : 'foo', 'b' : 'bar' }
foobar = namedtuple('foobar', d.keys())(**d)
print foobar

输出

Python 2.7.5 (default, May 30 2013, 16:55:57) [GCC] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> from collections import namedtuple
>>> d  =  { 'a' : 'foo', 'b' : 'bar' }
>>> foobar = namedtuple('foobar', d.keys())(**d)
>>> print foobar
foobar(a='foo', b='bar')
>>> 

Answer 4

class attrdict(dict):
    def __getattr__(self, key):
        return self[key]

o = attrdict(attr1='something', attr2=344)

o.attr1

但看起来你应该只使用标准字典。

Answer 5

我更喜欢 mwhite 的 dict 答案，但这是我过去如何使用 kwargs 的“魔法”（双关语意）来完成的。

class ObjectFromDict(object):
    def __init__(**kwargs):
        for k in kwargs:
            if k not in self.__dict__:
                setattr(k,v)

myObj = ObjectFromDict(**{'foo': 'bar', 'baz': 'monkey'})
print myObj.foo #bar